How do you find the general solution to 3y'+(y/x)=x

3y'+(y/x)=x

=> y' + (y/3x) = x/3 ............stanadard form

this is a linear first order differential equation, ( of the form y' + P(x)y = Q(x) )

here P(x) = 1/3x and Q(x) = x/3

Integrating factor( IF) = e^[integral(1/3x) dx ] ..................( IF = e^integral(P(x)dx) )

=> IF = e^(ln(x^1/3)) = x^(1/3)

solution is y * x^(1/3) = integral( (x/3) * x^(1/3) dx )....................solution is y*IF = integral(Q(x)*IFdx)

=> y * x^(1/3) = (1/3)* integral( (x)^(4/3) dx )

=> y * x^(1/3) = (1/3)* (3/7)* x^(7/3) + C

=> y * x^(1/3) = (1/7) * x^(7/3) + C

=> y = (1/7) * x^2 + Cx^(-1/3)

which is the solution,

hope it helped!!

Particular solution: Y = Ax^2

6Ax + Ax = x; A = 1/7

Homogeneous equation (separable):

3 y' + y/x = 0

3 dy/y = - dx/x

3 ln y = -- ln x + A

y^3 = B/x; y = C x^(-1/3)

General solution of inhomogeneous equation:

y = x^2 / 7 + C x^(-1/3)